I have Sphere with radius R. Inside the sphere I have Two circle. One circle is fixed defined by $\alpha$. Another circle can rotate and the orientation of that circle can be defined by $\beta$. From the centre of the sphere until the edge of the movable circle create $\alpha$ angle with $z$ axis. $h$ and $a$ are distance from the origin and base radius of this circle .enter image description here

Now when both circle intersect each other what will be the surface area of intersection? Using spherical coordinate system $r,\theta,\phi$, I know that my $\theta $ limit goes from $\gamma$ to $\alpha$. But I am not getting the limit for $\phi$ with this given parameters. I am trying to calculate the intersected surface area using only calculus and trigonometry for relevancy of my future problem.

  • $\begingroup$ @YuriyS I have edited the picture. Is it clear now? $\endgroup$ – T. an Jul 25 '19 at 8:18
  • $\begingroup$ To the edited version: you mean the area of a circular segment on the horizontal circle which is cut off by the inclined circle? $\endgroup$ – Yuriy S Jul 25 '19 at 8:20
  • $\begingroup$ Or the part of the sphere surface cut off by the two circles? This seems more likely $\endgroup$ – Yuriy S Jul 25 '19 at 8:21
  • $\begingroup$ @YuriyS It is the whole surface area in between two circle's intersection. $\endgroup$ – T. an Jul 25 '19 at 8:21
  • $\begingroup$ Yeah part of the sphere cut off by two circle. $\endgroup$ – T. an Jul 25 '19 at 8:22

I'll change the problem statement a little so I can use Cartesian coordinates at first.

Also let $R=1$, since we can always multiply the answer by $R^2$.

Let Circle 1 lie in the horizontal plane with elevation $d$ (Plane 1).

Let Circle 2 lie in the plane defined by the radius vector $\vec{h}$ (Plane 2).

Equation of Plane 1:


Equation of Plane 2:

$$h_x x+h_y y+h_z z=h^2$$

Equation of the sphere:


Now we transition to spherical coordinates:

$$x = r \sin \theta \cos \phi$$

$$y = r \sin \theta \sin \phi$$

$$z = r \cos \theta$$

Which gives us:


$$\cos \theta=d$$

$$h_x \sin \theta \cos \phi+h_y \sin \theta \sin \phi+h_z \cos \theta=h^2$$

We now have equations to determine the intersection points of the two circles on the sphere.

$$\pm \sqrt{1-d^2}(h_x \cos \phi+h_y \sin \phi)+h_z d=h^2$$

$$h_x \cos \phi+h_y \sin \phi= \pm\frac{h^2-h_z d}{\sqrt{1-d^2}}$$


$$h_x=\cos \delta, \qquad h_y= \sin \delta$$

Then we have:

$$\cos (\phi-\delta)=\pm\frac{h^2-h_z d}{\sqrt{1-d^2}}$$

$$\phi= \delta+\arccos \left(\pm \frac{h^2-h_z d}{\sqrt{1-d^2}} \right)$$

$$\phi= \arccos h_x+\arccos\left(\pm \frac{h^2-h_z d}{\sqrt{1-d^2}} \right)$$

$$\theta=\arccos d$$

We found the intersection point coordinates (note also the condition for their existence):

$$-1 \leq \frac{h^2-h_z d}{\sqrt{1-d^2}} \leq 1$$

I think now half of the problem is solved, since we can use those points to set up the limits of integration on the surface of the sphere.

  • $\begingroup$ Thanks for your great try...So for the $\phi$ another limit suppose to be $\pi-\phi$ right? $\endgroup$ – T. an Jul 25 '19 at 9:59
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    $\begingroup$ @T.an, not exactly, see my edit. I forgot that I have two solutions for $\sin \theta$ $\endgroup$ – Yuriy S Jul 25 '19 at 10:04
  • $\begingroup$ so one is for when circle goes up from right side and crossing each other and another is for left side when it makes reverse crossing? $\endgroup$ – T. an Jul 25 '19 at 10:07
  • $\begingroup$ @T.an, you make my head hurt with this question :). I'm better at algebra than geometry, so I hope you figure it out $\endgroup$ – Yuriy S Jul 25 '19 at 10:08
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    $\begingroup$ @T.an, important: my $\gamma$ is not the same as yours. I will edit $\endgroup$ – Yuriy S Jul 25 '19 at 10:36

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